If the Taylor Series of \ln(x) is known: \ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-... Can one find

namenerk 2021-12-25 Answered
If the Taylor Series of ln(x) is known:
ln(x)=(x1)12(x1)2+13(x1)214(x1)4+15(x1)5
Can one find the Taylor series of
f(x)=x1x2
by manipulating the Taylor series of ln(x) ?
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Expert Answer

Joseph Lewis
Answered 2021-12-26 Author has 43 answers
Make the substitution x=1u2 in order to obatin:
ln(1u2)=u2u42u63=k=1u2kk
Based on the ratio test, the corresponding function is analytic for |u|<1. Prescisely,
limk|ak+1ak|=limn(kk+1)|u2|=|u2|<1|u|<1
Hence we can differentiate the obtained expression in order to get:
2u1u2=2k=1u2k1f(x)=x1x2=k=1x2k1=x+x3+x5+
whenever |x|<1, and we are done.
Hopefully this helps!
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Paul Mitchell
Answered 2021-12-27 Author has 40 answers
user_27qwe
Answered 2021-12-30 Author has 229 answers

Notice that:
f(x)dx=x1x2dx=12duu=12ln|1x2|+C
(via the u-substitution u=1x2)
From your note:
ln(x)=(x1)12(x1)2+13(x1)2+...
So substitute 1x2 into x in the natural log taylor expansion, and take the derivative to obtain a series representation of f(x).

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